Examine the continuity of the function :
`f(x)={{:(x+1" , "xle2),(2x-1" , "xgt2):}` at x = 2.

Answer

To examine the continuity of the function

\[ f(x) = \begin{cases} x + 1 & \text{if } x \leq 2 \\ 2x - 1 & \text{if } x > 2 \end{cases} \]

at \( x = 2 \), we need to check the following conditions:

1. **Find \( f(2) \)**: This is the value of the function at \( x = 2 \).
2. **Find the left-hand limit \( f(2^-) \)**: This is the limit of the function as \( x \) approaches 2 from the left.
3. **Find the right-hand limit \( f(2^+) \)**: This is the limit of the function as \( x \) approaches 2 from the right.
4. **Check if \( f(2) = f(2^-) = f(2^+) \)**: If all these values are equal, then the function is continuous at \( x = 2 \).

### Step 1: Find \( f(2) \)

Since \( 2 \leq 2 \), we use the first case of the function:

\[ f(2) = 2 + 1 = 3 \]

### Step 2: Find \( f(2^-) \)

To find the left-hand limit, we consider values of \( x \) approaching 2 from the left (i.e., \( x \leq 2 \)):

\[ f(2^-) = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x + 1) = 2 + 1 = 3 \]

### Step 3: Find \( f(2^+) \)

To find the right-hand limit, we consider values of \( x \) approaching 2 from the right (i.e., \( x > 2 \)):

\[ f(2^+) = \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (2x - 1) = 2(2) - 1 = 4 - 1 = 3 \]

### Step 4: Check continuity

Now we check if \( f(2) = f(2^-) = f(2^+) \):

\[ f(2) = 3, \quad f(2^-) = 3, \quad f(2^+) = 3 \]

Since all three values are equal, we conclude that the function \( f(x) \) is continuous at \( x = 2 \).

### Final Conclusion

The function \( f(x) \) is continuous at \( x = 2 \).

---